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If I'll get an old HP calculator that has no working charger anymore, can I in that case use a new NON-HP charger that has the right voltage?

I have dealt with a place called International Calculator website www.internationalcalculator.com. They sell no longer made HP calculators (including the HP41C/CV/CX) and accessories such as battery chargers. I have purchased a working HP48SX, cards for it, and soft calculator cases from its manager, Don O'Rourke. He was very helpful. Also try Ebay www.ebay.com. I joined Ebay about two months ago and bid on a calculator book and later on an HP calculator, both of which I now have. Good luck!

Volts is volts... if you can get the proper voltage (mayby with a sutible current limiter) on the proper pins then it will work. This is easier said than done for some machines. Ebay or the HPMUSEUM classified ads are your best bets.

Not easily. The output voltage is not the whole story by any means, and HP chargers rarely give the specified voltage anyway.
For machines other than 'classics' (35, 45, 55, 65, 67, 70, 80) ,the charger is a special transformer. AC output. The output voltage of this transformer is designed to fall under load, in a specified way.
The 'classics' take a fairly complicated charger, which contians internal regulation circuitry... Oh, and don't get me started on the 'car' chargers. I've looked inside the one for the Woodstock series. 6 transsitors. And a complicated I-V output characteristic.
What machine are you getting that doesn't have a charger? Most HP chargers are still fairly easy to find second-hand, so you can probably get the right unit. Using a substitute might damage the calculator.

Hello everyone,

Just some thoughts prompted by Tony. I don't have access to any of these old calc's but I would imagine that it would be possible for anyone familiar with electronics to fathom the system used.

If the charger/power supply is doing the rectification and regulation, then a similar unit could be built. If a working system is available for 'probing' then that would be even better.

If the charger/power supply has no obvious rectification or regulation, then these must be on the calc's circuit board and again it would be possible to identify them. Also look for some form of cooling fin.

Regulators are usually ICs, but I suppose very old equipment might have a regulator built round an opamp or discrete transistors.

Go carefully with that multimeter...


Many of the HP chargers have been 'fathomed out' as you put it. The circuitry is often very simple, but it does depend on the characteristics of things like the transformer.
AFAIK, no HP handheld calculator series ever used 'traditional' voltage regulator chips. The earliest series (the 'classics') has a fairly complicated charger with internal regulation circuirty. Using discrete components. 4 diodes + an electrolytic cap to get about 16V DC from the transformer. Then a classic zener + emitter follower circuit to give a stabilised 4.2V (or so) for the logic. And 2 transistors (the base-emitter junction of one connected across a current sensing shunt resistor, the collector current of this trnansistor then controls the other one) as a constant current supply
to charge the NiCds. That charger could be reproduced I think as all the components are 'known'.
Some later series use the battery as a shunt regulator. The voltage across an NiCd under charge is not going to rise above 1.5V per cell. The 'charger' can be as simple a the transformer (in the AC adapter module) and a diode + resistor in series in the calculator. In fact the Woodstock and Spice series do just that.
The problem is that if the battery goes open-circuit then the terminal voltage rises and this tends to blow up logic chips :-(. In the topcats HP protected against this by a little circuuit that loads the output of the AC adapter if the battery terminal voltage tries to rise above the Vss (+6.2V) line from the logic board power converter circuit. That's why a Topcat with power converter problems will draw around 1A from the battery pack, BTW.
Anyway, these machines (Woodstocks, Spices, Topcats, HP41 printer, HPIL tape drive, etc (which use the Topcat battery, AC adapter and much the same charger circuit) depend on the characteristics of the transformer to work correctly. There's a linear falloff of AC output voltage with increasing load current (I've measured it). You can't use just any transformer and expect it to work (or indeed be safe for the calculator).
HP did sell adapters to charge some of these calculators from a 12V DC input (a car battery, for example). They're quite complicated inside (half a dozen transistors + R's and C's) and have a somewhat strange output characteristic. The one for the
Woodstock series (which I've had in bits on the bench) acts as a constant current source with an output current suitable for charging the NiCd pack until you load it more heavily. Then the output current jumps to a higher (but again constant on further increasing the load current). This second current is enough both to charge the battery and
run the calculator. If you had to make a charger for the Woodstocks (or the Spice series which have essentially the same battery and internal charger circuit) you might have to make something like the 12V input charger and run it from a bench PSU.

Tony, regarding the simpler AC adapters, you have enlightened me in earlier threads about the current limiting function of the resistance of the transformer secondary winding. If one were trying to substitute another transformer with the same open-circuit output voltage, and added an additional resistance in series so that the total resistance was the same as the original transformer secondary, could similar performance and protection of the calculator be expected?

Also, I understand that a transformer must be considered to transform resistance (or impedance) as well as voltage and current. This suggests the resistance or impedance of the primary winding would also act to limit current. Would adding a resistance in series with the primary as well as the secondary allow a transformer with the same open-circuit output voltage behave like another transformer (assuming the substitute had less resistance in both windings than the original)? Can this be generalized so that, for instance, a table of transformer winding DC resistances could be published which could be used to calculate series resistances to add to the measured resistance of a substitute transformer's windings?

In any case, I think the biggest problem with rigging up a substitute AC adapter is the varied and non-standard connectors HP used.

My guess is that you could use a normal transformer with a suitable resistor in series. I'd put it in series with the low-voltage secondary winding, but it should be possible to use one in series
with the primary instead. I'd not start by measuring the DC resistances of the transformer windings, though, as they'd not tell you the full story by any means.
What I'd do is take the results I'd measured from a real HP charger. Say it was 12V and 11 Ohms (those values are about right for a Topcat, btw). I'd take a 12V transformer and a 10 Ohm (say) resistor, link the resistor in series with the secondary and determine the intenral resistance
of the combination by doing the well-known (high school?) experiment of loading it with various resistors and measuring ther terminal voltage.
Then I'd adjust the resistor to give me the internal resistance I wanted. Would I trust that on a HP calculator? It should be OK, but...
As regards the connectors. the pins from disk drive power sockets will fit the pins on a Topcat charger connector. You'd just have to make the housing. And I suspect a cut-apart edge connector could be made to work in a Spice charger socket, but I've not tried it. I have made most types of male (as on the calculators) charger connector, which involved turning the pins from brass rod. Not trivial, but
possible in a reasonable home workshop.

I just read your earlier post re: HP67 charger. That's putting a 3421A to good use! I've got a 3421A with the 10 channel mux. I know I can use two of the channels as external switches. I've planned to do something similar to what you describe to experiment with rechargable batteries: charge them while monitoring voltage and current, then connect a load and monitor voltage and current over time to determine charge capacity. If you use the mux actuator channels to switch load currents, what precautions do you use to protect the relays on the mux board? I know they are rated 2A max but I think I want to use external relays so the mux will only have to switch the relay coil current. Recently I got a digital IO board for the 3421A (haven't tested it yet) which will probably be better to control the loads with external relays.

I didn't use the 'Actuator channels' to switch loads -- like you I was worried about damaging the relays. My 3421 has a 10 channel mux card and a digital I/O card in it.
I used the 8 outputs from the digital I/O card to control 8 external relays. I even stuck a ULN2803 chip between the digital outputs and the relay coils -- yes the digital I/O card should have been able to handle the coil current, but the chip was cheap and it removed a possible worry.
I used 7 of the relays to control a series chain of resistors (Each relay contact shorted out one resistor in the chain, resistors in an approximate doubling sequence, with a 10 Ohm in series with the whole lot so it couldn't totally short out the charger).
The last relay connected the resistor chain to the output of the charger. I used 2 of the analogue input channels of the 3421, one connected to the output of the charger, the other to the resistor chain.
The software does the following : With the 'last relay' open, it tries all 128 combiations of the other relays and measures the resistance of the chain. That way I can use chaep 5% wirewound resistors and only have to worry about calibrating the 3421, With the 'last relay' still open, measure the (AC) voltage at the terminals of the charger.
Then close the last relay and try all 128 combinations of the other relays again. Measure the voltage at the charger terminals, and calculate the current (V/R, where R is the known resistance of the chain for that relay setting). Then plot V against I for that device.
Incidentally, the reactive component of the internal impedance of the HP chargers seems to be negiligable, so it's quite OK just to use resistive loads and treat the thing like a DC problem.

Thanks for the tips! I like the idea of measuring the resistors instead of using precision ones, although I might use one low resistance precision resistor in series with the variable load to monitor the current with another voltmeter channel. I'm curious to know if you have anything to say about the 3421A voltmeter drifting up (I started another thread about this).

If you know the value of a resistor (which we've just agreed you can measure) and the voltage across it (the terminal voltage of the AC adapter) then you can
trivially work out the current. And for a DC circuit, or an AC one with negligable reactance (as here, amazingly) that's all you need to do.

I was going to say "because the resistance measurements are less accurate than the voltage measurements" but I thought I should check the specs and make sure. Now I wonder how the resistance measurement can be more accurate than the voltage considering the voltmeter function is involved in making the resistance measurement? The specs say the ohmmeter (on the lower ranges) is a little better than the voltmeter, except between 90 days and one year after calibration. The voltmeter is better on the 3V range than the 0.3V range, but even then the ohmmeter is equal to the voltmeter in accuracy. This seems to suggest that the current source used for the resistance measurement is perfect!

Could this be like Microsoft giving the Office people better info on the Windows API than outside developers? Did HP give the ohmmeter function a better voltmeter that it gave to the user via the front panel?